What is the maximum value of f(x) = -3x^2 + 12x - 5? (2019)

₹0.0

Question: What is the maximum value of f(x) = -3x^2 + 12x - 5? (2019)

Options:

Correct Answer: 7

Exam Year: 2019

Solution:

The vertex is at x = -12/(2*(-3)) = 2. The maximum value is f(2) = -3(2^2) + 12(2) - 5 = 7.

What is the maximum value of f(x) = -3x^2 + 12x - 5? (2019)

What is the maximum value of f(x) = -3x^2 + 12x - 5? (2019)

Step 1: Identify the function we are working with, which is f(x) = -3x^2 + 12x - 5.
Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c, where a = -3, b = 12, and c = -5.
Step 3: Determine the x-coordinate of the vertex using the formula x = -b/(2a). Here, b = 12 and a = -3.
Step 4: Calculate -b: -12.
Step 5: Calculate 2a: 2 * -3 = -6.
Step 6: Divide -12 by -6 to find the x-coordinate of the vertex: x = -12 / -6 = 2.
Step 7: Now, substitute x = 2 back into the function f(x) to find the maximum value.
Step 8: Calculate f(2): f(2) = -3(2^2) + 12(2) - 5.
Step 9: Calculate 2^2: 2^2 = 4.
Step 10: Multiply -3 by 4: -3 * 4 = -12.
Step 11: Multiply 12 by 2: 12 * 2 = 24.
Step 12: Now, combine the results: f(2) = -12 + 24 - 5.
Step 13: Calculate -12 + 24: 12.
Step 14: Finally, calculate 12 - 5: 12 - 5 = 7.
Step 15: The maximum value of the function f(x) is 7.

No concepts available.